Question

Assume that a company hires employees on Mondays, Tuesdays, or Wednesdays with equal likelihood.a. If two different employees are randomly​ selected, what is the probability that they were both hired on a Monday​?b. If two different employees are randomly​ selected, what is the probability that they were both hired on the same day of the week​?c. What is the probability that 7 people in the same department were all hired on the same day of the week​?d. Is such an event​ unlikely?

Answer 1

Answer:

$P(A) = \frac{1}{9}$

$P(B) = \frac{1}{3}$

$P(C) = \frac{1}{729}$

Step-by-step explanation:

We know that:

Only employees are hired during the first 3 days of the week with equal probability.

2 employees are selected at random.

So:

A. The probability that an employee has been hired on a Monday is:

$P(M) = \frac{1}{3}$.

If we call P(A) the probability that 2 employees have been hired on a Monday, then:

$P(A) =P(M\ and\ M)\\\\P(A)=( \frac{1}{3})(\frac{1}{3})\\\\P(A) = \frac{1}{9}$

B. We now look for the probability that two selected employees have been hired on the same day of the week.

The probability that both are hired on a Monday, for example, we know is $P(A) = \frac{1}{9}$. We also know that the probability of being hired on a Monday is equal to the probability of being hired on a Tuesday or on a Wednesday. But if both were hired on the same day, then it could be a Monday, a Tuesday or a Wednesday.

So

$P(B) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}\\\\P(B) = \frac{1}{3}$.

C. If the probability that two people have been hired on a specific day of the week is $3(\frac{1}{3}) ^ 2$, then the probability that 7 people have been hired on the same day is:

$P(C) = (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7\\\\P(C) = \frac{1}{729}$

D. The probability is $\frac{1}{729}$. This number is quite close to zero. Therefore it is an unlikely bastate event.